$|\operatorname{Aut}(K/F)| \leq [K:F]_s$ holds in general?

Let $K/F$ be finite extension. $$|\operatorname{Aut}(K/F)| \leq [K:F]$$ is a fundamental equation, and equality holds if the extension is Galois.

My question is; more strict inequality, $$|\operatorname{Aut}(K/F)| \leq [K:F]_s$$ holds in general? $[K:F]_s$ is separable degree of $K/F$. My text seems to be using this without assumption that $K/F$ is separable.

If there are counter examples, it is also appreciated. Thank you for your help.

Solution 1:

Lemma. Let $E/k$ be separable, and let $\sigma: k\rightarrow k^a$. Then $\sigma$ extends to an embedding of $E$ in $k^a$, and there are $[E:k]$ distinct extensions of $\sigma$. The number of such distinct monomorphism is $\lt [E:k]$ if $E/k$ is not separable.

Proof. We will prove the lemma by induction. Let $\alpha\in E$, then the minimal polynomial $\text{Irr}(\alpha,k)$ has no multiple roots. Any embedding of $k(\alpha)$ sends $\alpha$ to a root of $\text{Irr}(\alpha,k)$, so there are $\text{deg}(\text{Irr}(\alpha,k))=[k(\alpha):k]$ distinct extensions. For any $\beta\in k(\alpha)$, since $\text{Irr}(\beta,k(\alpha))\mid \text{Irr}(\beta,k)$, it follows from our assumption that $\text{Irr}(\beta,k(\alpha))$ is separable. By induction hypothesis, the number of distinct extensions of a given embedding of $k(\alpha)$ on $E$ is $[E:k(\alpha)]$. Hence the number of distinct extensions of $\sigma$ on $E$ is $[E:k(\alpha)]$. The second assertion is proven in essentially the same way.

Let $K$ be an extension of $F$. Given any $\alpha\in K$, let its minimal polynomial over $F$ be $f(x)$. Let the roots of $f(x)$ be $\alpha_1=\alpha, \alpha_2,\cdots, \alpha_ r$, then $f(x)=\prod_{j=1}^r (x-\alpha_j)^{m_j},$ and for each $\alpha_j$ there exists an isomorphism $\sigma: k(\alpha)\rightarrow k(\alpha_j)$, which may be extended to an automorphism of $K^a$. Then all $m_j$ are equal.

If $\text{char}F=0$, any irreducible polynomial in $F[x]$ is separable, so $[K:F]=[K:F]_s$. We assume that $\text{char}F=p\gt 0$. If $f(x)$ is inseparable over $F$, its formal derivative $f'(x)$ shares a common root with $f(x)$. But this is impossible unless $f'$ is identically $0$, and thus $p\mid m_j$ i.e. $f(x)=g(x^p)$ for some $g(x)\in F[x]$. Proceeding inductively, we have $f(x)=h(x^{p^m})$ for some separable polynomial $h(x)\in F[x]$. Then $h$ is irreducible in $F[x]$ and hence is the minimal polynomial of $\alpha^{p^m},$ and $\text{deg}(g)=r=[K:F]_s$. Therefore by our lemma above, given $\sigma: F\to F^a$, the number of distinct monomorphism extending $\sigma$ is $[K:F]_s$. Hence $|Aut(K/F)|\le [K:F]_s$.

We remark that if $K/F$ is normal, any monomorphism $\sigma: K\to F^a$ is an automorphism on $K$, so $|Aut(K/F)|= [K:F]_s$.